1/(1+x^2)の[0, t]での定積分がarctan(t)となる話
問
original: https://www.youtube.com/watch?v=PvL2RDp1H4Y
中心を点\(P\)に持つ円と直線\(g\)との交点\(Q\)を求める．
\(x\)座標を求める． $$\begin{eqnarray} x^2+\left(y+1\right)^2&=&1 \\x^2+\left(\color{red}{\frac{x}{t}-1}\color{black}{}+1\right)^2&=&1\;\cdots\;g:y=\frac{x}{t}-1を代入 \\x^2+\left(\frac{x}{t}\right)^2&=&1 \\x^2\left(1+\frac{1}{t^2}\right)&=&1 \\x^2\left(\frac{t^2+1}{t^2}\right)&=&1 \\x^2&=&\frac{t^2}{t^2+1} \\x&=&\sqrt{\frac{t^2}{t^2+1}}\;\cdots\;x\ge0 \\&=&\frac{t}{\sqrt{t^2+1}} \end{eqnarray}$$ \(y\)座標は\(g\)を用いて求める． $$\begin{eqnarray} y&=&\frac{x}{t}-1 \\&=&\frac{\frac{t}{\sqrt{t^2+1}}}{t}-1 \\&=&\frac{1}{\sqrt{t^2+1}}-1 \end{eqnarray}$$面積\(A_1\)(=四角形\(ABCO\))を求める．
$$\begin{eqnarray} A_1&=&\overline{OA}\cdot\overline{AB}\;\cdots\;長方形の面積 \\&=&t\cdot\frac{1}{2}\frac{1}{1+t^2} \\&=&\frac{t}{2\left(1+t^2\right)} \end{eqnarray}$$面積\(A_2\)(=凾数\(f\)と線分\(\overline{CB}\)及び\(y\)軸で囲まれる領域(図の\(x\lt0\)の領域の着色は間違い))を求める．
$$\begin{eqnarray} 2y&=&\frac{1}{1+x^2} \\1+x^2&=&\frac{1}{2y} \\x^2&=&\frac{1}{2y}-1 \\x^2&=&\frac{1}{2y}-1 \\&=&\frac{1-2y}{2y} \\x&=&\sqrt{\frac{1-2y}{2y}} \end{eqnarray}$$ $$\begin{eqnarray} A_2&=&\int_{\frac{1}{2\left(1+t^2\right)}}^{\frac{1}{2}} x\mathrm{d}y \\&=&\int_{\frac{1}{2\left(1+t^2\right)}}^{\frac{1}{2}} \sqrt{\frac{1-2y}{2y}} \mathrm{d}y \\&=&\int_{\sqrt{\frac{1}{\left(1+t^2\right)}}-1}^{0} \sqrt{\frac{1-\left(u+1\right)^2}{\cancel{2y}}}\cancel{\sqrt{2y}}\mathrm{d}u \\&&\;\cdots\;u=\sqrt{2y}-1,\;\frac{\mathrm{d}u}{\mathrm{d}y} =\frac{1}{2}\frac{1}{\sqrt{2y}}\cdot2=\frac{1}{\sqrt{2y}},\;\mathrm{d}y=\sqrt{2y}\mathrm{d}u \\&&\;\cdots\;\frac{1}{2}\rightarrow\sqrt{2\cdot \frac{1}{2}}-1=0 \\&&\;\cdots\;\frac{1}{2\left(1+t^2\right)}\rightarrow\sqrt{2\cdot \frac{1}{2\left(1+t^2\right)}}-1 =\frac{1}{\sqrt{1+t^2}}-1 \\&=&\int_{\frac{1}{\sqrt{1+t^2}}-1}^{0} \sqrt{1-\left(u+1\right)^2}\mathrm{d}u \end{eqnarray}$$面積\(B_1\)(=三角形\(PQR\))を求める．
$$\begin{eqnarray} B_1&=&\frac{1}{2}\overline{QR}\cdot\overline{PR}\;\cdots\;三角形の面積 \\&=&\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+1}}\cdot\left\{1+\left(\frac{1}{\sqrt{t^2+1}}-1\right)\right\} \\&=&\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+1}}\cdot\frac{1}{\sqrt{t^2+1}} \\&=&\frac{t}{2\left(1+t^2\right)} \end{eqnarray}$$面積\(B_2\)(=円\(x^2+(y-1)^2=1\)と線分\(\overline{QR}\)及び\(y\)軸で囲まれる領域)を求める．
$$\begin{eqnarray} x^2+\left(y+1\right)^2&=&1 \\x^2&=&1-\left(y+1\right)^2 \\x&=&\sqrt{1-\left(y+1\right)^2} \end{eqnarray}$$ $$\begin{eqnarray} B_2&=&\int_{\frac{1}{\sqrt{t^2+1}}-1}^{0} x \mathrm{d}y \\&=&\int_{\frac{1}{\sqrt{t^2+1}}-1}^{0} \sqrt{1-\left(y+1\right)^2} \mathrm{d}y \end{eqnarray}$$扇PQRの面積を求める．
$$\begin{eqnarray} 扇PQR&=&半径1の面積\cdot比率=\left(1\cdot1\cdot\pi\right)\cdot\frac{\theta}{2\pi} &=&\frac{\theta}{2} \end{eqnarray}$$\(\theta\)と\(t\)の関係式を求める．
$$\begin{eqnarray} \tan{\left(\theta\right)}&=&\frac{t-0}{0-(-1)}=\frac{t}{1}=t \\\tan^{-1}{\left(t\right)}&=&\theta \end{eqnarray}$$扇PQRの面積を介して．
$$\begin{eqnarray} 扇PQR&=&\frac{\theta}{2} \\&=&\frac{\tan^{-1}\left(t\right)}{2} \\&=&B_1+B_2 \\&=&A_1+A_2\;\cdots\;B_1=A_1,\;B_2=A_2 \\&=&\int_0^t \frac{1}{2}\frac{1}{1+x^2}\mathrm{d}x =凾数fの区間[0,t]における定積分 \end{eqnarray}$$ $$\begin{eqnarray} \int_0^t \frac{1}{1+x^2}\mathrm{d}x&=&\tan^{-1}\left(t\right) \end{eqnarray}$$n乗の3つの数の和を基本対称式で表す
n乗の3つの数の和を基本対称式で表す
\(\alpha, \beta, \gamma\)を解とする3次方程式を考える． $$\begin{eqnarray} \left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)&=&0 \\x^3-\left(\alpha+\beta+\gamma\right)x^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)x-\alpha\beta\gamma&=&0 \end{eqnarray}$$ \(\alpha,\beta,\gamma\)は式の解なので，\(x\)に代入しても式は成り立つ．代入した2つの式を得る． $$ \begin{eqnarray} \left\{ \begin{array}{l} \alpha^3-\left(\alpha+\beta+\gamma\right)\alpha^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha-\alpha\beta\gamma&=&0 \\\beta^3-\left(\alpha+\beta+\gamma\right)\beta^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta-\alpha\beta\gamma&=&0 \\\gamma^3-\left(\alpha+\beta+\gamma\right)\gamma^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma-\alpha\beta\gamma&=&0 \end{array} \right. \end{eqnarray} $$ 第一式には\(\alpha^{n-3}\)，第二式には\(\beta^{n-3}\)，第三式には\(\gamma^{n-3}\)を両辺に掛ける． $$ \begin{eqnarray} \left\{ \begin{array}{l} \alpha^{n-3}\left\{\alpha^3-\left(\alpha+\beta+\gamma\right)\alpha^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha-\alpha\beta\gamma\right\}&=&\alpha^{n-3}\cdot0 \\\beta^{n-3}\left\{\beta^3-\left(\alpha+\beta+\gamma\right)\beta^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta-\alpha\beta\gamma\right\}&=&\beta^{n-3}\cdot0 \\\gamma^{n-3}\left\{\gamma^3-\left(\alpha+\beta+\gamma\right)\gamma^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma-\alpha\beta\gamma\right\}&=&\gamma^{n-3}\cdot0 \end{array} \right. \\ \\\left\{ \begin{array}{l} \alpha^n&-\left(\alpha+\beta+\gamma\right)\alpha^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha^{n-2}&-\alpha\beta\gamma\alpha^{n-3}&=&0 \\\beta^n&-\left(\alpha+\beta+\gamma\right)\beta^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta^{n-2}&-\alpha\beta\gamma\beta^{n-3}&=&0 \\\gamma^n&-\left(\alpha+\beta+\gamma\right)\gamma^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma^{n-2}&-\alpha\beta\gamma\gamma^{n-3}&=&0 \end{array} \right. \end{eqnarray} $$ 3式を足し合わせ，n乗の和について解くことで．n乗の和を基本対称式で求める式を得る． $$\begin{eqnarray} &\alpha^n&-\left(\alpha+\beta+\gamma\right)\alpha^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha^{n-2}&-\alpha\beta\gamma\alpha^{n-3}&=&0 \\&\beta^n&-\left(\alpha+\beta+\gamma\right)\beta^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta^{n-2}&-\alpha\beta\gamma\beta^{n-3}&=&0 \\+)&\gamma^n&-\left(\alpha+\beta+\gamma\right)\gamma^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma^{n-2}&-\alpha\beta\gamma\gamma^{n-3}&=&0 \\\hline &\alpha^{n}+\beta^{n}+\gamma^{n} &-\left(\alpha+\beta+\gamma\right)\left(\alpha^{n-1}+\beta^{n-1}+\gamma^{n-1}\right) &+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\left(\alpha^{n-2}+\beta^{n-2}+\gamma^{n-2}\right) &-\alpha\beta\gamma\left(\alpha^{n-3}+\beta^{n-3}+\gamma^{n-3}\right)&=&0 \end{eqnarray}$$ $$\begin{eqnarray} \alpha^{n}+\beta^{n}+\gamma^{n}&=& \left(\alpha+\beta+\gamma\right)\left(\alpha^{n-1}+\beta^{n-1}+\gamma^{n-1}\right) -\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\left(\alpha^{n-2}+\beta^{n-2}+\gamma^{n-2}\right) +\alpha\beta\gamma\left(\alpha^{n-3}+\beta^{n-3}+\gamma^{n-3}\right) \end{eqnarray}$$\(\alpha, \beta, \gamma\)は任意の数でよいのでそれを\(x, y, z\)とすれば以下の式を得る． $$\begin{eqnarray} x^{n}+y^{n}+z^{n}&=&\left(x+y+z\right)\left(x^{n-1}+y^{n-1}+z^{n-1}\right)-\left(xy+yz+xz\right)\left(x^{n-2}+y^{n-2}+z^{n-2}\right)+xyz\left(x^{n-3}+y^{n-3}+z^{n-3}\right) \end{eqnarray}$$
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